# How do you prove that the function f(x)=(x-5)/ |x-5| is continuous everywhere but x=5?

Nov 1, 2015

Use the definition of $\left\mid u \right\mid = \left\{\begin{matrix}- u & \text{if" & u < 0 \\ u & "if} & u \ge 0\end{matrix}\right.$ to rewrite $f$ as a piecewise function.

#### Explanation:

$f \left(x\right) = \left\{\begin{matrix}- 1 & \text{if" & x<5 \\ 1 & "if} & x > 5\end{matrix}\right.$

So, $f$ is constant on $\left(- \infty , 5\right)$, hence continuous on $\left(- \infty , 5\right)$

and $f$ is constant on $\left(5 , \infty\right)$, hence continuous on $\left(5 , \infty\right)$.

$f \left(5\right)$ is not defined, so $f$ is not continuous at $5$.